ExaminerPrep › IAL Physics (Edexcel) › Flashcards
IAL Physics (Edexcel) · Topic 2.10 · 35 cards
Resistance, temperature and light flashcards
Test yourself one card at a time: say the answer, flip, then mark it. Cards you know drop out; the rest come back until you know them. Your progress stays in this browser.
- NTC thermistorA semiconductor resistor whose resistance decreases as its temperature increases (negative temperature coefficient).
- Light-dependent resistor (LDR)A semiconductor resistor whose resistance decreases as the intensity of the light falling on it increases.
- Drift velocityThe small average velocity of the charge carriers along a conductor, superimposed on their fast random motion.
- Charge carrier number density, nThe number of free charge carriers (e.g. conduction electrons) per unit volume of the material; unit m⁻³.
- ResistanceR = V/I: the potential difference across a component divided by the current in it, at that point.
- Current in terms of charge carriersI = nqvA (n = carriers per m³, q = charge per carrier, v = drift velocity, A = cross-sectional area). (Yes)
- Resistance from a point on an I–V graphR = V/I at that point (not the gradient of the curve). (Yes)
- How does R depend on n and v for a given conductor at fixed p.d.?R ∝ 1/(nv), from I = nqvA and R = V/I. (No — derive it)
- Output p.d. of a sensor potential dividerV_out = V_in × R_out ÷ (R₁ + R₂), where R_out is the resistance the output is taken across. (No — derive it)
- Sketch: R against θ for a metalAlmost straight line, positive gradient, positive intercept at 0 °C; not through the origin.
- Sketch: R against θ for an NTC thermistorCurve falling steeply at low temperature, levelling off at high temperature; never reaches zero.
- Sketch: R against light intensity for an LDRCurve falling steeply at low intensity and levelling off at high intensity; very high (≈ MΩ) in the dark.
- Sketch: I–V graph of a filament lampThrough the origin and symmetrical about it (the same shape, inverted, in the third quadrant); gradient decreasing as |V| increases (curves towards the V axis).
- Circuit symbol for a thermistorResistor box with a diagonal line through it ending in a short horizontal foot.
- Circuit symbol for an LDRResistor box inside a circle with two arrows pointing in towards it (light falling on it).
- Why does a metal's resistance increase with temperature?Ions vibrate with larger amplitude → electrons collide with them more often → drift velocity falls; n unchanged → I falls → R rises.
- Why does an NTC thermistor's resistance decrease with temperature?More charge carriers are released, so n rises greatly; this outweighs the small fall in drift velocity from larger vibrations → I rises → R falls.
- Why does an LDR's resistance decrease in brighter light?More light energy absorbed → more electrons released as charge carriers → n increases (v unchanged) → R falls.
- Why is a filament lamp's I–V graph curved?Larger current → hotter filament → larger lattice vibrations → higher resistance, so each extra volt gives less extra current.
- Why is n effectively constant when a metal is heated?Each metal atom has already released its conduction electrons; heating releases no significant extra carriers.
- Why does a current heat a conductor?Charge carriers transfer energy to the lattice ions in collisions, increasing the ions' vibrations.
- Temperature sensor: where is the output taken, and why?Across the fixed resistor (thermistor on top): when hotter, the thermistor's share of the total resistance falls, so the fixed resistor's p.d. rises.
- Dusk switch: where is the output taken, and why?Across the LDR: when darker, its resistance rises, its share of the supply p.d. rises, so the output rises.
- Measuring R against temperature: key precautionsWater bath, stir, wait for steady readings, keep connections dry, use a small current (ohmmeter) so the component does not self-heat.
- 3.0 A in a copper wire of area 1.0 mm²; n = 8.5 × 10²⁸ m⁻³. Drift velocity?v = I/(nqA) = 3.0 ÷ (8.5 × 10²⁸ × 1.60 × 10⁻¹⁹ × 1.0 × 10⁻⁶) = 2.2 × 10⁻⁴ m s⁻¹.
- Metal wire at constant p.d.: current falls from 2.00 A to 1.50 A as it warms. Ratio of new to old drift velocity?n, q, A constant, so v ∝ I: 1.50 ÷ 2.00 = 0.75.
- A lamp takes 0.25 A at 1.5 V and 1.0 A at 12 V. Resistance at each?1.5 ÷ 0.25 = 6.0 Ω; 12 ÷ 1.0 = 12 Ω — higher when the filament is hotter.
- 9.0 V across a thermistor (8.0 kΩ) in series with 4.0 kΩ. P.d. across the 4.0 kΩ resistor?9.0 × 4.0 ÷ 12.0 = 3.0 V.
- 12 V dusk switch: output across the LDR must be 9.0 V when the LDR is 30 kΩ. Fixed resistor?Fixed resistor has 3.0 V; R ÷ 30 = 3.0 ÷ 9.0, so R = 10 kΩ.
- Semiconductor heated: n rises × 5.0, drift velocity falls by 10%, constant p.d. Factor change in R?I changes by 5.0 × 0.90 = 4.5, so R changes by 1/4.5 = 0.22.
- Which factor in I = nqvA changes: metal, thermistor, LDR?Metal: v falls (n constant). Thermistor: n rises greatly (beats a small fall in v). LDR: n rises (v unchanged).
- “The thermistor's ions vibrate less when hot”Wrong: every lattice vibrates more when hotter. The thermistor's R falls because n rises.
- Resistance of a lamp from its I–V graphRead one point and use R = V/I; do not use the gradient of the tangent.
- Six-mark explanation structureLinked chain: cause → vibrations or carriers → collisions/drift velocity or n → current → resistance. One point per link.
- Deciding which way a sensor output movesAsk whether the component the output is across takes a larger or smaller share of the total resistance.